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Does anyone know of a simple proof that the general linear group GL(N) does not admit spinor representations?

Thank you!

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Could you clarify what you mean? If $V$ is the $2n$-dimensional vector space with bilinear form $x_1 y_2 + x_2 y_2 + \cdots x_n y_n$, and $SO(V)$ is the group of symmetries of $V$ preserving this quadratic form, then $GL_n$ maps to $SO(V)$ by $g \mapsto \left( \begin{smallmatrix} g & 0 \\ 0 & g^{-T} \end{smallmatrix} \right)$ and this map lifts to the double cover of $SO(V)$. So $GL_n$ acts on spinors for $2n$-dimensional space. (If we are working over the reals, this is embedding $GL_n(\mathbb{R})$ into $Spin(n,n)$.) I imagine there is a good question here, but it needs some explanation. –  David Speyer Feb 12 '13 at 16:24
    
i believe the missing word here is faithful. take a look at lemma 5.23 in lawson-michelsohn. –  johndoe Feb 12 '13 at 17:20
    
Hi, thanks for your reply. I hope I can clarify the question. I read the following statement in Green, Schwartz, Witten-Superstring theory, Vol 2, page 272: "Spinors form a representation of SO(N) which does not arise from a representation of GL(N,R)." If I understand your comment correctly, what you are saying is that one can think of the action of GL(N,R) on a spinor by restricting GL(N,R) to its SO(N) subgroup, but I think the question is whether the spinor retains its transformation properties under the whole group. This seems to be impossible. Do you know why? Thanks! –  Nuno Feb 12 '13 at 17:34
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Spinors aren't a representation of $SO(N)$, they are a representation of the double cover $Spin(N)$ of $SO(N)$. Perhaps the question is why there is no corresponding double cover of $GL(N)$, extending the action of $Spin(N)$? –  David Speyer Feb 12 '13 at 17:39
    
Yes, that is precisely the issue. Do you have any comments/ideas? –  Nuno Feb 12 '13 at 23:20
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1 Answer 1

up vote 13 down vote accepted

The statement in GSW that you quote has to be interpreted properly. When they write, "Spinors form a representation of $\mathrm{SO}(N)$ which does not arise from a representation of $\mathrm{GL}(N,\mathbb{R})$", they really mean either "Spinors form a representation of ${\frak{so}}(N)$ which does not arise from a representation of ${\frak{gl}}(N,\mathbb{R})$" (i.e., it's really a statement about representations of Lie algebras) or else they mean what some sources call "multi-valued representations", i.e., representations of the simply-connected covers of the groups $\mathrm{SO}(N)$ and $\mathrm{GL}(N,\mathbb{R})$. Moreover, 'arise from' means that there is no (finite dimensional) representation of ${\frak{gl}}(N,\mathbb{R})$ that, when restricted to ${\frak{so}}(N)\subset{\frak{gl}}(N,\mathbb{R})$, contains a copy of a 'spinor representation' of ${\frak{so}}(N)$.

Either way, what it comes down to is this: If $\pi:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{SL}(n,\mathbb{R})$ is the nontrivial double cover of $\mathrm{SL}(n,\mathbb{R})$, then any finite-dimensional representation $\hat\rho:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{R})$ must factor through $\pi$, i.e., $\hat\rho = \rho\circ\pi$ for some (unique) representation $\rho:\mathrm{SL}(n,\mathbb{R})\to\mathrm{GL}(k,\mathbb{R})$.

The reason for this is topological: Although $\pi_1\bigl(\mathrm{SL}(n,\mathbb{R}),I_n\bigr)$ is nontrivial for $n\ge 2$ (being $\mathbb{Z}$ when $n=2$ and $\mathbb{Z}_2$ when $n>2$), the group $\pi_1\bigl(\mathrm{SL}(n,\mathbb{C}),I_n\bigr)$ is trivial for all $n\ge 2$. Using this fact, the proof goes like this:

If $\hat\rho:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{R})$ is any representation, let $\hat\rho':{\frak{sl}}(n,\mathbb{R})\to {\frak{gl}}(k,\mathbb{R})$ denote the induced Lie algebra homomorphism. This complexifies to a Lie algebra homomorphism $(\hat\rho')^{\mathbb{C}}:{\frak{sl}}(n,\mathbb{C})\to {\frak{gl}}(k,\mathbb{C})$ and, since $\mathrm{SL}(n,\mathbb{C})$ is simply-connected, this is induced by a Lie group homomorphism $\rho^{\mathbb{C}}:\mathrm{SL}(n,\mathbb{C})\to \mathrm{GL}(k,\mathbb{C})$, which restricts to a Lie group homomorphism $\rho:\mathrm{SL}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{C})$ with associated Lie algebra homomorphism $\rho':{\frak{sl}}(n,\mathbb{R})\to{\frak{gl}}(k,\mathbb{C})$. By construction, this homomorphism $\rho'$ must be the composition of $\hat\rho':{\frak{sl}}(n,\mathbb{R})\to {\frak{gl}}(k,\mathbb{R})$ with the inclusion ${\frak{gl}}(k,\mathbb{R})\hookrightarrow {\frak{gl}}(k,\mathbb{C})$. In particular, $\rho'$ maps into ${\frak{gl}}(k,\mathbb{R})$ after all. Now it's clear that $\hat\rho = \rho\circ\pi$.

In particular, if $\mathrm{Spin}(n)\subset \hat{\mathrm{SL}}(n,\mathbb{R})$ is the pre-image under $\pi$ of $\mathrm{SO}(n)\subset \mathrm{SL}(n,\mathbb{R})$, then any finite dimensional representation of $\hat{\mathrm{SL}}(n,\mathbb{R})$ restricts to a representation of $\mathrm{Spin}(n)$ that factors through $\mathrm{SO}(n)$ and hence cannot be (or even contain) a spinor representation, since those do not factor through $\mathrm{SO}(n)$.

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Thank you very much Robert for your answer. –  Nuno Feb 17 '13 at 1:14
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